Replace the epsilon and delta in the definition of continuity, what kind of continuity do we get?

Asked Mar 21 '19 at 06:53

Active Mar 21 '19 at 07:26

Viewed 91 times

1) $f : X \to Y$ is a map. For all $x_0\in X$, if for every $\delta > 0$, there is $\varepsilon > 0$ such that $d_X(x_0, x) < \delta$ whenever $d_Y(f(x_0), f(x)) < \varepsilon$.

2)$f : X \to Y$ is a map. For all $x_0\in X$, if for every $ \varepsilon> 0$, there is $\delta > 0$ such that $d_Y(f(x_0), f(x)) < \delta$ whenever $d_X(x_0, x) < \varepsilon$.

Is map $f$ continuous in the two cases? If not, what kind of continuity is this?

For the second the case, I guess this definition is much stronger than continuity?

edited Mar 21 '19 at 07:26

asked Mar 21 '19 at 06:53

High GPA

3,776
15
44

1

You have shown that $f^{-1}$ is continuous at $f(x_0)$ (if it exists at all) instead of showing that $f$ is continuous at $x_0.$ – little o Mar 21 '19 at 06:57
1

The function is bounded in the second case – vidyarthi Mar 21 '19 at 07:24
What is the context where you have found such a weird looking questions? – little o Mar 21 '19 at 07:27
1

@Dbchatto67 Due to my humble love of math – High GPA Mar 21 '19 at 07:27
1

why the downvote? – vidyarthi Mar 21 '19 at 07:28
1

see here – vidyarthi Mar 21 '19 at 07:30
Every bounded function is continuous according to definition 2). So you cannot call it a form of continuity. – Kavi Rama Murthy Mar 21 '19 at 07:42

Replace the epsilon and delta in the definition of continuity, what kind of continuity do we get?

0 Answers0